TRANSLATOR’S NOTE
In preparing this translation, my assumption has been that the interested audience likely
consists of both mathematicians and historians of mathematics. To satisfy the latter, I have
attempted, as near as possible, to mimic Euler’s phrasing, and especially his mathematical
notation, with a few exceptions in cases where Euler’s notation might be confusing to a
modern reader. These include:
• Use of “arcsin, arccos” rather than “A sin, A cos” for the inverse trigonometric functions
• Use of, for example, “cos2 u” rather than “cos u2 ” to denote the square of the cosine;
• Clarifying the argument of the square root operator in paragraphs 42 and 51.
In most other cases, I have attempted to copy Euler’s original notation.
My parenthetical notes are enclosed in brackets; material found in parentheses is found
so in the original texts.
George Heine
[email protected]
December 2013
Principles of Spherical Trigonometry
Drawn from the Method of the Maxima and
Minima
Leonhard Euler
(G. Heine, tr.)
December 16, 2014
Since one knows that the arcs of great circles, drawn on the surface of a
sphere, represent the shortest path from one point to another, a spherical
triangle could be defined as follows: given three points on the surface of a
sphere, let a spherical triangle be the space enclosed between these three
points. Thus, since the sides of a spherical triangle are the shortest lines
which can be drawn from one angle to another, the method of maxima and
minima could be used to determine the sides of a spherical triangle and from
this could be found the relations which exist between the angles and sides,
which is exactly the content of spherical Trigonometry. For the three points,
where are found the angles, will determine the three sides as well as the three
angles, and these six things will always have such a relationship among them,
that any three being known, one can from this determine the three others.
This, therefore, is a property which the spherical triangles have in common
with plane triangles, which are the subject of elementary Trigonometry. For,
just as a plane triangle is the space enclosed between three points marked on
a plane, when the shortest path is drawn from one to another, which is, on
the plane, a straight line, so a spherical triangle is the space enclosed between
three points marked on the surface of a sphere, when these three points are
joined by the shortest lines that can be drawn on the same surface. Now,
it is clear that a spherical triangle changes into a planar triangle, when the
radius of a sphere becomes infinitely large, so that a planar surface can be
regarded as the surface of an infinitely large sphere.
1
No doubt, one will object that it goes against the rules of the method,
to want to use the calculus of the maxima and minima to establish the
foundations of spherical Trigonometry; it seems useless, moreover, to derive
these again from other priniciples, since those which have been used until
now, are founded on elementary Geometry, whose rigor serves as a rule for
all the other parts of Mathematics. But I remark, first, that the method
of maxima and minima thereby acquires somewhat of a new luster, when I
show that by itself it leads to the resolution of spherical triangles; moreover,
it is always useful to arrive by different routes at the same truth, since our
spirits will not fail thereby to arrive at new understandings.
But I also may argue that the method of maxima and minima is much
more general than the ordinary method. For the latter is limited to triangles
formed on plane or spherical surfaces, while the former extends to any surface
whatever. Thus, if one asks the nature of triangles formed on spheroidal or
conic surfaces, whose sides are the shortest lines that can be drawn from
one angle to the other, the ordinary method would not be suitable to such
research; it would be absolutely necessary to resort to the method of maxima
and minima, without which one would not even be in a state to know the
shortest lines, which form the sides of these triangles.
From this, it is understood that this research could well become of great
importance; for the surface of the Earth is not spherical, but spheroidal;
a triangle formed on the surface of the Earth belongs to the sort of which
I just mentioned. To see this, one only needs to imagine three points on
the surface of the Earth which are joined by the shortest path which leads
from one to the other, or formed by a cord stretched from one to the other;
for it is thus that those triangles must be represented, which are used in
the operations for the measure of the Earth. Certainly, it is true that such
triangles are ordinarily regarded as plane and rectilinear, and it is quite
accurate, when one calculates on the basis of spherical triangles. But if one
succeeds in making these triangles much larger, and one wants to calculate
with the greatest possible precision, it would without doubt be necessary to
investigate the true nature of such triangles, which cannot be fully known
without using the method of the maxima and minima.
Having, therefore, seen the importance of this method in the subject which
it concerns, it will do no harm to apply this method to the resolution of
spherical triangles, since on one hand, this inquiry will serve as a basis and a
model for the resolution of triangles formed on a arbitrary spheroidal surface;
on the other hand it will furnish us with considerable explanations as much for
2
O
M
θ
A ζ
m
n
B
y
x
P p
Figure 1: Euler’s original figure, labels for ζ and θ added
spherical Trigonometry itself as for the method of the maxima and minima,
from which one will know more and better its extent and great usage. For,
once one has shown that most mechanical and physical problems are resolved
very quickly by means of this method, it will only be very pleasant to see
that the same method brings such a great help to the resolution of problems
in pure Geometry.
To begin this inquiry in such a manner that it applies equally to the sphere
and to any spheroid whatever, I first consider two opposite points on the
sphere as its poles, and the great circle equally distant from them will represent the equator, and the shortest lines drawn from a pole to each point
on the equator will represent the meridians which are perpendicular to the
equator. Now on the sphere, any side of a spherical triangle can be regarded
as part of the equator, and when it is a right triangle, one of the sides which
forms the right angle can always be supposed to be a portion of the meridian,
since the two poles can be chosen freely. But it will not be the same when
the surface is not spherical, but spheroidal. However, I will only speak here
of spherical surfaces, reserving spheroidal surfaces for another Memoire.
Problem 1
1. Given (Fig. 1) the arc AP on the equator AB, and the arc P M on a
meridian OP , find, on the spherical surface, the shortest line AM , which
can be drawn from point A to point M .
3
Solution
Setting the half-diameter of the sphere = 1, let
the arc of the equator AP = x,
and
the arc of the meridian P M = y.
In addition, let
the sought-after arc AM = s;
and let it be prolonged an infinitely small distance to m, so that M m = ds,
and let the meridian Omp be drawn through m, so that the element M n is
perpendicular to Omp. From this, one will have P p = dx and mn = dy, and
since P p is to M n as the sine of the total meridian OP =1 to the sine of the
arc OM , or to the cosine of P M = y, one will have M n = dx cos y, and the
triangle M nm to n, being right, gives
p
M m = ds = dy 2 + dx2 cos2 y
and therefore
Z p
AM = s =
dy 2 + dx2 cos2 y .
Thus, the problem is to find such a relation between x and y, that if known
values such as AP and P M are given, the value of the integral
Z p
dy 2 + dx2 cos2 y
becomes as small as possible. To this effect, set dy = p dx, in order to reduce
this integral to the form
Z
p
dx pp + cos2 y ;
R
and since I have demonstrated1 that when the integral formula Z dx, where
Z is a function of x, y, and p, with dZ = M dx + N dy + P dp, must become
as small or large as possible, this happens by the equation
N dx − dP = 0.
1
See, for example, Proposition III, Chapter II, from E65(1743) – Methodus inveniendi
lineas curvas maximi minimive proprietate gaudentes, sive solutio problematis isoperimetrici lattissimo sensu accepti. [tr]
4
Thus, making application to our case, we shall have
p
Z = pp + cos2 y;
thus
p dp
dy sin y cos y
+p
;
dZ = − p
pp + cos2 y
pp + cos2 y
and consequently
M = 0,
sin y cos y
N = −p
,
pp + cos2 y
and
p
P =p
.
pp + cos2 y
Now, since M = 0 and therefore dZ = N dy + P dp, we multiply the equation
N dx − dP = 0 by p, which, since dy = p dx, will become N dy − p dP = 0,
or N dy = p dP , and, this value being substituted for N dy will give
dZ = p dP + P dp ,
whose integral is Z = P p + C, or
p
pp + cos2 y = p
pp
pp + cos2 y
+ C;
which reduces to
cos2 y = C
p
pp + cos2 y;
from which is elicited
CC pp = cos2 y(cos2 y − CC),
or
p
cos y cos2 y − CC
dy
p=
=
.
dx
C
Thus the relation between x and y is expressed in the separated differential
equation2
dx =
2
C dy
p
cos y cos2 y − CC
“équation différentielle séparée” [tr]
5
and from this is obtained
p
dx cos2 y
;
ds = dx pp + cos2 y =
C
thus
dy cos y
ds = p
,
cos2 y − CC
and the arc itself
Z
s=
dy cos y
p
.
cos2 y − CC
Corollary 1
2. Thus, the equation
dx =
C dy
p
cos y cos2 y − CC
expresses the nature of the line AM , which has the property, that taking any
portion whatever, this will be the shortest line that can be drawn between
its endpoints on the surface of the sphere. Now, I have shown elsewhere that
this line is also a great circle of the sphere; here it does not matter to our
purpose, what relation this line has with the sphere, provided that we know
it is the shortest between its endpoints.
6
Corollary 2
3. Having found
dx =
C dy
p
cos y cos2 y − CC
we shall have
C dy
M n = dx cos y = p
.
cos2 y − CC
Mn
expresses the tangent of the angle AM P and consequently we
Now
mn
have:
C
tang AM P = p
.
cos2 y − CC
Furthermore, having
dy cos y
M m = ds = p
,
cos2 y − CC
Mn
expresses the sine of the angle AM P , so that
Mm
p
cos2 y − CC
C
and cos AM P =
.
sin AM P =
cos y
cos y
the fraction
Corollary 3
4. Moreover, setting y = 0, the point M will arrive at A and then the fraction
dy
dy
dx
will express the tangent of the angle P AM , and
its sine and
its
dx
ds
ds
cosine. Now having cos y = 1,
C dy
dx = √
1 − CC
and ds = √
7
dy
,
1 − CC
from which we conclude
√
1 − CC
,
tang P AM =
C
sin P AM =
√
1 − CC,
and
cos P AM = C.
Corollary 4
5. Thus, if we introduce this angle P AM in place of the constant C, and set
P AM = ζ, we shall have, since C = cos ζ, the two following equations:
dx =
dy cos ζ
p
cos y cos2 y − cos2 ζ
and ds = p
dy cos y
cos2 y − cos2 ζ
.
Moreover, if we name the angle AM P = θ, we shall have
cos ζ
tang θ = p
,
cos2 y − cos2 ζ
cos ζ
sin θ =
,
cos y
and
p
cos2 y − cos2 ζ
cos θ =
.
cos y
Corollary 5
6. It yet remains to integrate the two differential equations which express
the values of dx and ds. By integration, it is found that
x = arcsin
C sin y
√
,
cos y 1 − CC
C sin y
cos ζ sin y
√
=
sin ζ cos y
cos y 1 − CC
or
sin x =
or
p
p
cos2 y − CC
cos2 y − cos2 ζ
cos s = √
=
sin ζ
1 − CC
and
p
cos2 y − CC
s = arccos √
,
1 − CC
Corollary 6
7. Here, then, are the quantities ζ and y, and the other quantities x, s, θ so
8
determined:
cos ζ sin y
, cos x =
sin x =
sin ζ cos y
p
cos2 y − cos2 ζ
cos ζ sin y
, and tang x = p
sin ζ cos y
cos2 y − cos2 ζ
sin y
sin s =
,
sin ζ
p
cos2 y − cos2 ζ
cos ζ
, and tang s = p
sin ζ
cos2 y − cos2 ζ
p
cos2 y − cos2 ζ
cos ζ
, and tang θ = p
cos y
cos2 y − cos2 ζ
cos ζ
sin θ =
,
cos y
cos s =
cos θ =
Corollary 7
8. There is only one irrational formula,
p
cos2 y − cos2 ζ ,
in the equations we have found; by eliminating it, we shall obtain:
cos θ
cos θ
cos s
= cos y,
= sin ζ,
cos x
cos x
cos s
tang x
tang x
tang s
= cos ζ,
= sin y,
tang s
tang θ
tang θ
cos ζ sin y
sin y
sin x =
, cos x tang s =
, cos x tang θ
sin ζ cos y
sin ζ cos y
cos ζ sin y
sin y
cos s tang x =
,
sin s =
,
cos s tang θ
sin ζ
sin ζ
cos ζ sin y
sin y
cos θ tang x =
, cos θ tang s =
,
sin θ
cos y
cos y
=
=
=
=
=
sin ζ
,
cos y
sin y
,
cos ζ
cos ζ
,
sin ζ cos y
cos ζ
,
sin ζ
cos ζ
.
cos y
Corollary 8
9. Having here five quantities, x, y, s, ζ, and θ, which form part of the right
spherical triangle AP M , we take from the equalities just found those which
9
Figure 2
contain three of the quantities, and reduce them to a simpler form:
I.
II.
III.
IV.
V.
cos s = cos x cos y,
cos θ = sin ζ cos x,
tang x = cos ζ tang s,
tang x = sin y tang θ,
tang y = sin x tang ζ;
VI. sin y = sin ζ sin s,
VII. cos s tang ζ tang θ = 1,
VIII. tang y = cos θ tang s,
IX. cos ζ = sin θ cos y,
whence, given any two quantities, one can find from them the three others,
without the need to extract any roots, provided that one adds here this tenth
one:
X. sin x = sin θ sin s ,
which follows immediately from the three first formulae on the left in paragraph 7.
Problem 2
10. Exhibit the rules for the resolution of all cases of right spherical triangles.
Solution
Let the angles (Fig. 2) be marked by A, B, C, with C being the right angle,
and the sides by the lower-case letters a,b,c, corresponding to their opposite
10
angles, so that c is the hypotenuse and a and b the supports. Then comparing
this triangle with the previous figure, we shall have
s = c,
x = b,
y = a,
ζ = A,
and θ = B.
Now everything turns on the fact that, given two of these five quantities, the
three others are determined from them; then the formulae reported will yield
the following resolutions for all possible cases:
The two
given quantities
I. a, b
II.
a, c
III. b, c
IV.
a, A
V. a, B
VI.
b, A
VII.
b, B
Determination of the three others
tang a
,
sin b
sin a
cos c
,
sin A =
,
cos b =
cos a
sin c
tang b
cos c
,
cos A =
,
cos a =
cos b
tang c
sin a
tang a
,
sin c =
,
sin b =
tang A
sin A
tang a
tang b = sin a tang B, tang c =
,
cos B
tang b
tang a = sin b tang A, tang c =
,
cos A
sin b
tang b
sin a =
,
sin c =
,
tang B
sin B
cos c = cos a · cos b,
VIII. c, A
sin a = sin c sin A,
c, B
sin b = sin c sin B,
IX.
X. A, B
cos a =
cos A
,
sin B
11
tang A =
tang b
sin a
tang a
cos B =
tang c
sin b
sin B =
sin c
cos A
sin B =
,
cos a
tang B =
cos A = cos a sin B
cos B = cos b sin A
sin A =
cos B
cos b
1
cos c tang A
1
tang a = tang c cos B, tang A =
cos c tang B
cos B
1
cos b =
,
cos c =
sin A
tang A tang B
tang b = tang c cos A,
tang B =
Corollary 1
11. From the above it is evident that side a with its opposite angle A enters
into these formulae, exactly as does side b with its opposite angle B, so that
it makes no difference which of the two sides a and b one wishes to take for
the base, exactly as the nature of the subject requires.
Corollary 2
12. The large number of formulae which express the relationship between
the various parts of the right triangle, are reduced to the following formulae,
whose number is smaller; it suffices to learn these by heart.
sin b
sin a
or sin c =
I. sin c =
sin A
sin B
II.
cos c = cos a cos b
III.
cos c = cot A cot B
tang b
tang c
cos B
V. sin A =
cos b
tang b
VI. sin a =
tang B
IV.
cos A =
tang a
tang c
cos A
or sin B =
cos a
tang a
or sin b =
tang A
or
cos B =
Corollary 3
13. It is only necessary to note these six formulae, which contain many of the
properties of right spherical triangles, and it will be possible to solve every
imaginable case of such triangles.
Problem 3
14. Find the area of a right spherical triangle.
Solution
In the right triangle AP M (Fig. 1) let the base AP = x and the side P M = y,
12
and having drawn the infinitely close meridian Omp, one will have P p = dx
and mn = dy. Moreover, having M n = dx cos y, the element of the area
P M mp will be = dx dy cos y, taking dx for a constant. Thus, the area itself
will
P M mp = dx sin y ,
which, being the differential of the area of the triangle AP M , this will be
Z
=
dx sin y .
Now we have found that
dx =
dy cos ζ
p
,
cos y cos2 y − cos2 ζ
where ζ marks the angle P AM ; consequently, the area of the triangle will be
Z
dy sin y cos ζ
p
=
.
cos y cos2 y − cos2 ζ
In place of y we introduce the angle AM P = θ; and because
p
cos2 y − cos2 ζ
cos ζ
sin θ =
and cos θ =
,
cos y
cos y
we shall have
dθ cos θ =
dy cos ζ sin y
;
cos2 y
thus
dθ =
dy cos ζ sin y
p
,
cos y cos2 y − cos2 ζ
so that the sought-after area of the triangle becomes
Z
= dθ = θ + Const .
In order to assign this constant its proper value, one must consider that the
area should vanish when the point M collapses into A, in which case the
13
O
a
E
y
x
α
s
φ
M
n
m
B
A
Figure 3: Copy of Euler’s figure; labels for a,x,y,s,α, and φ added
angle θ becomes 90◦ − ζ; thus it must be that 90◦ − ζ + Const = 0, so that
Const = ζ − 90◦ . Consequently, the sought-after area of the triangle will be
= ζ + θ − 90◦ ;
in other words, the excess of the sum of the two angles ζ and θ over a right
angle shall express the area of the triangle AP M .
Corollary 1
15. Thus, the sum of the angles P AM and AM P is always greater than a
right angle, and the excess becomes larger as the area of the triangle increases.
And the product of the great-circle arc measuring this excess, multipied by
the radius of the sphere, will give the area of the spherical triangle.
Corollary 2
16. From this, the area of an arbitrary spherical triangle is easily deduced:
for, since such a triangle can be resolved into two right triangles, one will
find its area, when one multiplies the excess over 180◦ of the sum of the three
angles, by the radius of the sphere.
Problem 4
17. Given two arbitrary points E and M on the surface of a sphere(Fig. 3),
find the shortest line EM between them.
14
Solution
From one of the poles O, let meridians OE and OM be drawn, where the
location of the latter is regarded as variable. We denote
the meridian OE = a,
OM = x,
and the angle EOM = y.
Morever, let the desired quantities be
the arc EM = s,
the angle OEM = α,
and the angle OM E = φ,
the last being variable with the quantities x, y, and s, while a and α remain
constant. Let the infinitely close meridian Om be drawn, to which is drawn
from M the perpendicular M n. One will have
mn = dx,
the angle M Om = dy,
and M n = dy sin x,
where we take the radius of the sphere to be one. Then we shall have
tang φ =
dy sin x
Mn
=
,
mn
dx
or
sin φ =
Now having ds =
dy sin x
ds
and
cos φ =
dx
.
dy
p
dx2 + dy 2 sin2 x, it is required that the formula
Z q
dx2 + dy 2 sin2 x
be a minimum. To this end, we set dy = p dx; the formula to be minimized
becomes
Z
Z
q
2
dx 1 + pp sin x = Z dx,
so that
Z=
q
1 + pp sin2 x.
15
Now in general, if one has dZ = M dx+N dy+P dp, the equation to minimize3
is N dx − dP = 0; applied to the present case, we have
N =0
and
p sin2 x
P =p
.
1 + pp sin2 x
But since N = 0, our equation will be
dP = 0
P = Const.,
and thus
Consequently, we shall have
p sin2 x
p
= C,
1 + pp sin2 x
or
dy sin2 x
p
=C;
dx2 + dy 2 sin2 x
that is,
dy sin2 x
= sin x sin φ = C .
ds
To evaluate the constant C, we must consider that, when making the angle
EOM = y vanish, x becomes = a and φ = 180◦ − α, or sin φ = sin α, so in
this case we have sin a · sin α = C. Consequently, the nature of the minimum
yields the equation
dy sin2 x
p
= sin a sin α.
dx2 + dy 2 sin2 x
But it is still necessary to integrate the differential equation; writing C again
in place of sin a sin α, we have
dy =
C dx
sin x
p
sin2 x − CC
,
dy sin2 x
and since ds =
, we have
C
dx sin x
ds = p
.
sin2 x − CC
3
Another reference to Euler’s work on the calculation of variations. See the footnote
to paragraph 1[tr].
16
Now one finds through the rules of integration that
C cos x
C cos a
√
√
+ arcsin
sin x 1 − CC
sin a 1 − CC
p
p
sin2 x − CC
sin2 a − CC
√
√
= − arccos
+ arccos
,
sin x 1 − CC
sin a 1 − CC
p
p
sin2 x − CC
sin2 a − CC
s = − arccos √
+ arccos √
1 − CC
1 − CC
cos x
cos a
= − arcsin √
+ arcsin √
,
1 − CC
1 − CC
y = − arcsin
where the added constants are such that, making y = 0 and s = 0, x becomes
= a. But the two arcs of circles being reduced to one will give
p
p
C cos a sin2 x − CC − C cos x sin2 a − CC
,
y = arcsin
(1 − CC) sin a sin x
p
p
cos a sin2 x − CC − cos x sin2 a − CC
s = arcsin
,
1 − CC
from which we derive the following two equations:
p
p
(1 − CC) sin a sin x sin y = C cos a sin2 x − CC − C cos x sin2 a − CC,
p
p
(1 − CC) sin s = cos a sin2 x − CC − cos x sin2 a − CC .
But, taking the cosines of the angles y and s, we shall have:
q
(1 − CC) sin a sin x cos y = (sin2 a − CC)(sin2 x − CC) + CC cos a cos x
q
(1 − CC) cos s = (sin2 a − CC)(sin2 x − CC) + cos a cos x.
And replacing C by its value sin a sin α, since
p
sin2 a − CC = − sin a cos α,
for we regard here the angle α as obtuse, so that the angle φ is acute from
the point E; because, setting y = 0, the angle φ becomes 180◦ − α, thus its
17
cosine is − cos α. We shall have:
p
sin α cos a sin2 x − sin2 a sin2 α + sin α cos α sin a cos x,
p
(1 − sin2 a sin2 α) sin x cos y = − cos α sin2 x − sin2 a sin2 α + sin a cos a sin2 α cos x,
p
(1 − sin2 a sin2 α) sin s = cos a sin2 x − sin2 a sin2 α + sin a cos α cos x,
p
(1 − sin2 a sin2 α) cos s = − sin a cos α sin2 x − sin2 a sin2 α + cos a cos x,
(1 − sin2 a sin2 α) sin x sin y =
to which must be added
sin x sin φ = sin a sin α .
Corollary 1
18. Since sin a sin α = sin x sin φ, one will have:
p
sin2 x − sin2 a sin α = + sin x cos φ.
Thus, our four formulae become:
(I)
(II)
(III)
(IV)
(1 − CC) sin y = sin α cos a cos φ + cos α cos x sin φ
(1 − CC) cos y = − cos α cos φ + sin α cos a cos x sin φ
(1 − CC) sin s = cos a sin x cos φ + sin a cos α cos x
(1 − CC) cos s = − sin a cos α sin x cos φ + cos a cos x,
setting, to abbreviate, CC in place of sin2 a sin2 α or sin2 x sin2 φ.
Corollary 2
19. These four formulae can be combined in several different ways, from
which simpler formulae can be deduced. First, let us take
(I) · cos α + (II) · sin α cos a,
and we shall have
(1 − CC)(cos α sin y + sin α cos a cos y) = (cos2 α + sin2 α cos2 a) cos x sin φ.
18
Now
cos2 α + sin2 α cos2 a = 1 − sin2 α sin2 a = 1 − CC,
from which we derive
cos α sin y + sin α cos a cos y = cos x sin φ ==
sin a sin α
,
tang x
or
tang x sin y + tang α tang x cos a cos y = tang α sin a.
Corollary 3
20. Let make this combination
(I) · sin α cos a − (II) · cos α,
resulting in
(1 − CC)(sin α cos a sin y − cos α cos y) = (sin2 α cos2 a + cos2 α) cos φ
= (1 − CC) cos φ,
from which, dividing by 1 − CC, we get
sin α cos a sin y − cos α cos y = cos φ.
Corollary 4
21. The combination
(I) · sin x − (III) · sin α
gives
(1 − CC)(sin x sin y − sin α sin s) = 0 ,
or
sin x sin y = sin α sin s,
19
whence, since sin x sin φ = sin a sin α,
sin a sin y = sin φ sin s ,
in other words, the proportion
sin a : sin φ = sin x : sin α = sin s : sin y
Corollary 5
22. This combination:
(I) · sin a cos α sin x + (IV) · sin α cos a
gives
(1 − CC)(sin a cos α sin x sin y + sin α cos a cos s)
= cos x(sin a cos2 α sin x sin φ + sin α cos2 a).
Now, since
sin x sin φ = sin a sin α, the value of the formula becomes
sin α cos x(sin2 a cos2 α + cos2 a) = (1 − sin2 a sin2 α) sin α cos x = (1 − CC) sin α cos x,
so that, dividing by 1 − CC, one will obtain
sin a cos α sin x sin y + sin α cos a cos s = sin α cos x,
which becomes, since sin y =
sin α sin s
,
sin x
sin a cos α sin s + cos a cos s = cos x
Corollary 6
23. Now this combination
(I) · cos a − (IV) · cos α sin φ
20
gives
(1 − CC)(cos a sin y − cos α sin φ cos s) = cos φ(sin α cos2 a + sin a cos2 α sin α sin x sin φ) ,
whose value is, since sin x sin φ = sin a sin α,
sin α cos φ(cos2 a + sin2 a cos2 α) = (1 − CC) sin α cos φ ;
thus, dividing by 1 − CC, one has
cos a sin y − cos α sin φ cos s = sin α cos φ ,
which, since sin y =
sin φ sin s
, changes into
sin a
cos a sin φ sin s − sin a cos α sin φ cos s = sin a sin α cos φ ,
or
tang φ sin s − cos α tang a tang φ cos s = sin α tang a .
Corollary 7
24. We consider the combination
(II) · cos a sin x − (III) · cos α .
which gives
(1 − CC)(cos a sin x cos y + cos α sin s) = cos x(sin α cos2 a sin x sin φ + sin a cos2 α) ,
whose value, since sin x sin φ = sin a sin α, will be
sin a cos x(sin2 α cos2 a + cos2 α) = (1 − CC) sin a cos x .
Then dividing through by 1 − CC, one will have
cos a sin x cos y + cos α sin s = sin a cos x ,
21
and since sin s =
sin x sin y
, one will obtain
sin α
sin α cos a sin x cos y + cos α sin x sin y = sin α sin a cos x ,
or
tang α cos a tang x cos y + tang x sin y = tang α sin a ,
just as in paragraph 19.
Corollary 8
25. This combination
(II) · sin a cos α − (III) · cos a sin α sin φ
gives
(1 − CC) (cos a sin α sin s sin φ − sin a cos α cos y) = cos φ sin a cos2 α + cos2 a sin α sin x sin φ .
whose value, since sin x sin φ = sin a sin α, is
cos a cos φ cos2 α + cos2 a sin2 α = (1 − CC) sin a cos φ .
Thus, dividing by 1 − CC, one will have
cos a sin α sin s sin φ − sin a cos α cos y = sin a cos φ.
Now, having sin s =
sin a sin y
, one will obtain
sin φ
cos a sin α sin y − cos α cos y = cos φ,
just as in paragraph 20.
Corollary 9
26. Now this combination
(II) · sin a sin x − (IV) · 1
gives
(1 − CC)(sin a sin x cos y − cos s) = cos a cos x(sin a sin α sin x sin φ − 1) ,
22
which since sin x sin φ = sin a sin α, has the value
cos a cos x sin2 a sin2 α − 1 = −(1 − CC) cos a cos x .
Thus, dividing by −(1 − CC), one will have
cos s − sin a sin x cos y = cos a cos x .
Corollary 10
27. This combination:
(II) · 1 − (IV) · sin α sin φ
gives
(1 − CC) (cos y − sin α sin φ cos s) = cos α cos φ (sin a sin α sin x sin φ − 1) ,
so that
sin α sin φ cos s − cos y = cos α cos φ .
Corollary 11
28. The combination:
(III) · sin a cos α + (IV) · cos a
gives
(1 − CC) (sin a cos α sin s + cos a cos s) = cos x sin2 a cos2 α + cos2 a ;
thus
sin a cos α sin s + cos a cos s = cos x,
just as in paragraph 22.
23
Figure 4
Corollary 12
29. Finally, the combination
(III) · cos a − (IV) · sin a cos α
gives
(1 − CC) (cos a sin s − sin a cos α cos s) = sin x cos φ cos2 a + sin2 a cos2 α ,
thus
cos a sin s − sin a cos α cos s = sin x cos φ =
sin a sin α cos φ
,
sin φ
or
tang φ sin s − tang a tang φ cos α cos s = tang a sin α ,
just as in paragraph 23.
Problem V
30. Find all the properties (Fig. 4) between the sides and the angles of an
arbitrary spherical triangle.
24
Solution
Whatever be the proposed spherical triangle ABC, one of the angles A can
be regarded as the pole of the sphere; then the sides AB and AC will each
be meridians, and the third side BC the shortest line that can be drawn on
the surface of the sphere from point B to point C, so that this triangle ABC
can be compared with the figure ECM , which we have just considered in the
previous problem. Thus, if we use the letters A, B, C to denote the angles
of the same name, and if we set the sides AB = c, AC = b, and BC = a, the
previous designations are reduced to the present ones in this fashion:
Former designations. . . . . . . . . . . . .
New designations . . . . . . . . . . . . . . .
a, x, s, y , α , φ
c, b , a, A, B, C
The formulae found in the corollaries of the preceding problem will now
furnish us the following properties for the spherical triangle ABC:
I.
sin a : sin A = sin b : sin B = sin c : sin C
by paragraph 21
II.
cos C = cos c sin A sin B − cos A cos B
cos B = cos b sin A sin C − cos A cos C
cos A = cos a sin B sin C − cos B cos C
by paragraph 20
by analogy
by paragraph 27
cos c = cos C sin a sin b + cos a cos b
III. cos b = cos B sin a sin c + cos a cos c
cos a = cos A sin b sin c + cos b cos c
IV.
by analogy
by paragraph 22
by paragraph 26
sin a tang C − sin B tang c = cos a cos B tang C tang c by paragraph 23
sin b tang A − sin C tang a = cos b cos C tang A tang a by analogy
sin c tang B − sin A tang b = cos c cos A tang B tang b by paragraph 19
And it is to these four properties, which all the formulae that we found in
the previous problem are reduced.
Corollary 1
31. The first property contains a well-known quality of all spherical triangles,
by which we know that the sines of the sides have the same ratio as the sines
of their opposite angles.
25
Corollary 2
32. Thus, in a spherical triangle, if we know one side with its opposite angle,
and besides this another angle, or side, we shall immediately find the side,
or the angle, opposite.
Corollary 3
33. Each of the formulae which we have just found contains only four quantities belonging to the triangle. Therefore, if three of these are known, the
fourth may be determined.
Corollary 4
34. Therefore, from these [formulae] one can derive the rules for the resolution of all spherical triangles. Now as there are six objects in each triangle,
namely, the three sides and the three angles, if three of these are known, the
other three can be found, as we shall see in the following problems.
Problem 6
35. Given the three sides of a spherical triangle (Fig. 4), find the angles.
Solution
In the spherical triangle ABC let the three sides AB = c, AC = b, and
BC = a be given. It is required to find the three angles A, B, and C; this is
done by means of the third property, which furnishes us:
cos a − cos b cos c
,
sin b sin c
cos b − cos a cos c
cos B =
,
sin a sin c
cos c − cos a cos b
cos C =
.
sin a sin b
cos A =
26
Corollary 1
36. Thus we have
1 − cos A =
sin b sin c + cos b cos c − cos a
,
sin b sin c
or
1 − cos A =
cos(b − c) − cos a
,
sin b sin c
because
cos(b − c) = cos b cos c + sin b sin c.
Corollary 2
37. Now, since in general
1
1
cos p − cos q = 2 sin (q − p) sin (p + q) ,
2
2
our formula will change into
1 − cos A =
2 sin 12 (a − b + c) sin 21 (a + b − c)
.
sin b sin c
Thus, since
1
1 − cos A = 2(sin A)2 ,
2
we shall have
s
1
sin A =
2
sin 12 (a − b + c) sin 21 (a + b − c)
,
sin b sin c
27
and in the same way
s
sin 12 (b − a + c) sin 21 (b + a − c)
,
sin a sin c
s
sin 12 (c − a + b) sin 12 (c + a − b)
.
sin a sin b
1
sin B =
2
1
sin C =
2
Corollary 3
38. In adding unity to the cosines found, one has
1 + cos A =
cos a − cos b cos c + sin b sin c
cos a − cos(b + c)
=
.
sin b sin c
sin b sin c
Thus, since 1 + cos A = 2(cos 21 A)2 , the same conversion will give
s
sin 12 (b + c − a) sin 21 (b + c + a)
sin b sin c
s
sin 12 (a + c − b) sin 21 (a + c + b)
sin a sin c
s
sin 12 (a + b − c) sin 21 (a + b + c)
sin a sin b
1
cos A =
2
1
cos B =
2
1
cos C =
2
28
Corollary 4
39. From these are derived the tangents of the half-angles A, B, C:
s
sin 12 (a − b + c) sin 12 (a + b − c)
1
tang A =
2
sin 12 (b + c − a) sin 12 (b + c + a)
s
sin 12 (b − a + c) sin 12 (b + a − c)
1
tang B =
2
sin 12 (a + c − b) sin 12 (a + c + b)
s
sin 12 (c − a + b) sin 12 (c + a − b)
1
tang C =
2
sin 12 (a + b − c) sin 12 (a + b + c)
Corollary 5
40. These formulae are very useful for performing the computation by means
of logarithms. Now, having found one of the angles, say A, the other two are
found very easily; by the first property one has
sin c sin A
sin b sin A
and
sin C =
,
sin a
sin a
provided that it is known whether the angles are larger or smaller than a
right angle. But, in using the formulae found, this ambiguity disappears,
since one finds the half-angles, which are always smaller than a right angle.
sin B =
Corollary 6
41. The tangents of the half-angles provide other significant formulae, because, multiplying two together, one will have:
sin 12 (a + b − c)
1
1
tang A tang B =
,
2
2
sin 12 (a + b + c)
and since
1
1
sin p + sin q = 2 sin (p + q) cos (p − q)
2
2
29
and
1
1
sin p − sin q = 2 sin (p − q) cos (p + q)
2
2
one will obtain
2 sin 12 (a + b) cos 12 c
1
1
1 + tang A tang B =
2
2
sin 21 (a + b + c)
and
2 sin 12 c cos 12 (a + b)
1
1
.
1 − tang A tang B =
2
2
sin 21 (a + b + c)
Corollary 7
42. Likewise, in adding or subtracting two of these tangents, one obtains
q 1
1
1
sin
(a
+
c
−
b)
±
sin
(b
+
c
−
a)
sin 2 (a + b − c)
1
1
2
2
q
q
tang A ± tang B = q
2
2
sin 1 (b + c − a) sin 1 (a + c − b) sin 1 (a + b + c)
2
2
2
or
sin 12 (a + c − b) ± sin 12 (b + c − a)
1
1
,
tang A ± tang B =
2
2
tang 12 C sin 12 (a + b + c)
if one introduces the value of the tangent of C. Thus, using the reduction
shown above [paragraph 41], we shall have these two equations:
2 sin 12 c cos 12 (a − b)
1
1
tang A + tang B =
2
2
tang 12 C sin 12 (a + b + c)
2 sin 12 (a − b) cos 12 c
1
1
tang A − tang B =
.
2
2
tang 12 C sin 12 (a + b + c)
30
Corollary 8
43. Now, since
tang 12 A + tang 12 B
1
,
tang (A + B) =
2
1 − tang 21 A tang 12 B
we shall find by the formulae of the two previous corollaries:
cos 21 (a − b)
1
tang (A + B) =
2
tang 21 C cos 12 (a + b)
and similarly
cos 21 (a − c)
1
tang (A + C) =
2
tang 12 B cos 21 (a + c)
cos 21 (b − c)
1
.
tang (B + C) =
2
tang 12 A cos 12 (b + c)
Corollary 9
44. In the same way, since
tang 12 A − tang 21 B
1
tang (A − B) =
,
2
1 + tang 12 A tang 12 B
we shall have:
sin 12 (a − b)
1
tang (A − B) =
2
tang 21 C sin 12 (a + b)
and similarly
sin 12 (a − c)
1
tang (A − C) =
2
tang 21 B sin 12 (a + c)
sin 21 (b − c)
1
tang (B − C) =
.
2
tang 21 A sin 12 (b + c)
31
Problem 7
45. In a spherical triangle, being given the three angles, find the three sides.
Solution
Let ABC be a spherical triangle, whose angles A, B, and C are given. It is
required to find the three sides
AB = c,
AC = b,
and
BC = a.
Now property (II) of paragraph 30 gives us the cosines of these sides expressed
as follows:
cos A + cos B cos C
sin B sin C
cos B + cos A cos C
cos b =
sin A sin C
cos C + cos A cos B
cos c =
.
sin A sin B
cos a =
Corollary 1
46. From this we derive from this the following formulae:
cos A + cos(B + C)
,
sin B sin C
cos A + cos(B − C)
1 + cos a =
.
sin B sin C
1 − cos a = −
Now, since in general cos p + cos q = 2 cos 21 (p + q) cos 12 (p − q), we shall have
2 cos 21 (A + B + C) cos 21 (B + C − A)
sin B sin C
1
2 cos 2 (A + B − C) cos 12 (A − B + C)
1 + cos a =
.
sin B sin C
1 − cos a = −
32
Corollary 2
47. Thus, since 1 − cos a = 2(sin 12 a)2 and 1 + cos a = 2(cos 12 a)2 , we obtain
the following formulae:
s
cos 12 (A + B + C) cos 21 (B + C − A)
1
sin a = −
2
sin B sin C
s
cos 12 (A + B + C) cos 21 (A + C − B)
1
sin b = −
2
sin A sin C
s
cos 12 (A + B + C) cos 21 (A + B − C)
1
,
sin c = −
2
sin A sin B
where it must be noted, that since the sum of the angles A + B + C is always
greater that two right angles, the half-sum is greater than a right angle, and
therefore its cosine is negative.
Corollary 3
48. For the cosines of the half-sides one will have:
s
cos 12 (A + B − C) cos 21 (A − B + C)
1
cos a =
2
sin B sin C
s
cos 12 (B + A − C) cos 21 (B − A + C)
1
cos b =
2
sin A sin C
s
cos 12 (C + A − B) cos 21 (C − A + B)
1
cos c =
,
2
sin A sin B
and these formulae facilitate the use of logarithms.
Corollary 4
49. From the sine and cosine of the half-sides their tangents are easily de-
33
rived; these will be
1
tang a =
2
s
1
tang b =
2
s
1
tang c =
2
s
−
−
−
cos 12 (A + B + C) cos 21 (B + C − A)
,
cos 21 (A + B − C) cos 12 (A − B + C)
cos 12 (A + B + C) cos 21 (A + C − B)
,
cos 21 (B + A − C) cos 12 (B − A + C)
cos 12 (A + B + C) cos 21 (A + B − C)
,
cos 21 (C + A − B) cos 12 (C − A + B)
where again, one can easily use the calculus of logarithms.
Corollary 5
50. Multiplying two of these tangents together, one will obtain
cos 12 (A + B + C)
1
1
tang a tang b = −
.
2
2
cos 12 (A + B − C)
Now from this the two following formulae are derived:
2 cos 12 (A + B) cos 21 C
1
1
1 − tang a tang b =
,
2
2
cos 21 (A + B − C)
2 sin 12 C sin 21 (A + B)
1
1
.
1 + tang a tang b =
2
2
cos 12 (A + B − C)
Corollary 6
51. And if we add or subtract together two of these formulae, we shall obtain:
q
1
1
cos
(B
+
C
−
A)
±
cos
(A
+
C
−
B)
− cos 12 (A + B + C)
1
1
2
2
q
q
tang a ± tang b = q
,
2
2
1
1
1
cos (A + B − C) cos (A + C − B) cos (B + C − A)
2
2
34
2
Now having
1
tang c =
2
s
−
cos 12 (A + B + C) cos 21 (A + B − C)
,
cos 12 (C + A − B) cos 21 (C − A + B)
one will have
cos 12 (B + C − A) ± cos 21 (A + C − B) tang 21 c
1
1
.
tang a ± tang b =
2
2
cos 12 (A + B − C)
Corollary 7
52. From this is obtained, by the simplifications shown:
2 cos 12 C cos 12 (A − B) tang 12 c
1
1
tang a + tang b =
2
2
cos 12 (A + B − C)
and
2 sin 12 (A − B) sin 21 C tang 12 c
1
1
tang a − tang b =
.
2
2
cos 12 (A + B − C)
Corollary 8
53. Thus we find as above:
cos 12 (A − B)
1
1
tang (a + b) =
tang c
1
2
2
cos 2 (A + B)
cos 21 (A − C)
1
1
tang (a + c) =
tang b
1
2
2
cos 2 (A + C)
cos 21 (B − C)
1
1
tang (b + c) =
tang a .
1
2
2
cos 2 (B + C)
35
Corollary 9
54. In the same way, the tangents of the half-difference of the sides will be:
sin 12 (A − B)
1
1
tang c
tang (a − b) =
1
2
2
sin 2 (A + B)
sin 12 (A − C)
1
1
tang (a − c) =
tang b
1
2
2
sin 2 (A + C)
sin 21 (B − C)
1
1
tang (b − c) =
tang a .
1
2
2
cos 2 (B + C)
The use of these formulas will be of great importance in the problems which
follow.
Problem 8
55. In a spherical triangle (Fig. 4), given two sides with the angle between
them, find the third side and the other two angles
Solution
Let ABC be a triangle, for which are given the two sides AB = c and AC = b,
together with the angle A between them, and it is required to find the side
BC = a and the angles B and C.
To begin, the third formula of the third property [paragraph 30] yields
cos a = cos A sin b sin c + cos b cos c,
and the third formula of the fourth property gives the angle B,
tang B =
sin A tang b
sin c − tang b cos c cos A
from which one takes by analogy:
tang C =
sin A tang c
.
sin b − tang c cos b cos A
36
Now the expressions for the cotangents will be more convenient, so that one
has the following formulae for the solution:
cos a = cos A sin b sin c + cos b cos c
sin c cot b − cos c cos A
cot B =
sin A
sin b cot c − cos b cos A
cot C =
.
sin A
Corollary 1
56. Since
cos b cos c =
1
1
cos(b − c) + cos(b + c)
2
2
sin b sin c =
1
1
cos(b − c) − cos(b + c),
2
2
and
the cosine of side a can be expressed by the addition and subtraction of
simple cosines in this manner:
cos a =
1
1
1
cos(A − b + c) + cos(A + b − c) − cos(A − b − c)
4
4
4
1
1
1
− cos(A + b + c) + cos(b − c) + cos(b + c) .
4
2
2
Corollary 2
57. But if one wishes to use logarithms, this formula is less convenient. However, logarithms could be applied here by introducing a new angle u, setting
tang u =
cos A sin b
,
cos b
37
or tang u = cos A tang b, and having found this angle u, one will have:
cos a = tang u cos b sin c + cos b cos c =
cos b cos(c − u)
,
cos u
from which the side a will easily be found by means of logarithms.
Corollary 3
58. The same introduction of the angle u, so that
tang u = cos A tang b,
also makes the other formulae appropriate for the application of logarithms,
for one will have:
tang B =
sin A tang b cos u
tang A sin u
sin A tang b
=
=
.
sin c − tang u cos c
sin(c − u)
sin(c − u)
As for the other angle C, it will be found by the rule
sin C =
cos A sin c
.
sin a
Corollary 4
59. But the most convenient search for the angles B and C will be drawn
from the formulae given in paragraphs 43 and 44, whence one will have
cos 21 (b − c)
1
1
cot A ,
tang (B + C) =
1
2
2
cos 2 (b + c)
sin 12 (b − c)
1
1
tang (B − C) =
cot A .
1
2
2
sin 2 (b + c)
For having half of the sum along with half of the difference, one will have
each separately, and from this the side a can then be obtained by the rule
sin a =
sin b
sin c
sin A =
sin A .
sin B
sin C
38
Problem 9
60. In a spherical triangle (Fig. 4), being given two angles with the side between them, find the third angle with the two sides.
Solution
Let ABC be a triangle, in which are given the two angles A and B, with the
side between them AB = c; it is required to find the third angle C together
with the two other sides AC = b and BC = a.
To begin, the first formula of the second property(paragraph 30) gives:
cos C = cos c sin A sin B − cos A cos B
and the third formula of the fourth property gives
tang b =
sin c tang B
,
sin A + cos c cos A tang B
tang a =
sin c tang A
.
sin B + cos c cos B tang A
and by analogy
From which, taking the cotangents, one will have the following solution:
cos C = cos c sin A sin B − cos A cos B
cot A sin B + cos c cos B
cot a =
sin c
cot B sin A + cos c cos A
cot b =
.
sin c
Corollary 1
61. The two sides will be found more easily from the formulae of paragraphs
39
52 and 53, from which one takes:
cos 12 (A − B)
1
1
tang c
tang (a + b) =
1
2
2
cos 2 (A + B)
sin 12 (A − B)
1
1
tang (a − b) =
tang c ,
1
2
2
sin 2 (A + B)
where it is easy to use logarithms.
Corollary 2
62. After having found the sides a and b, one will easily find the angle C,
since it is
sin A
sin B
sin C =
sin c =
sin c ,
sin a
sin b
or one could also, if desired, express cos C by simple cosines in this fashion:
cos C =
1
1
1
cos(c + A − B) + cos(c − A + B) − cos(c − A − B)
4
4
4
1
1
1
− cos(c + A + B) − cos(A − B) + cos(A + B) .
4
2
2
Problem X
63. In a spherical triangle, given two sides with an angle not between them
(Fig. 4) or equivalently, given two angles with a side not between them, find
the other quantities belonging to the triangle.
Solution
Let ABC be the triangle in which is given, for the first case, the two sides
BC = a and AC = b together with the angle A; one immediately knows the
angle B because
sin B =
sin A
sin b .
sin a
40
For the other case, let A and B be the given angles, together with the side
BC = a, one will immediately have the side b because
sin b =
sin a
sin B .
sin A
Consequently, in the one and the other case, one will be able to regard as
given the two sides BC = a and AC = b, as well as the two opposite angles
A and B. Thus, it is a matter of finding the side AB = c and the angle C.
Now the first formula of the fourth property gives:
sin a tang C − sin B tang c = cos a cos B tang C tang c ,
from which, transposing the sides a and b with the angles A and B, we shall
have
sin b tang C − sin A tang c = cos b cos A tang C tang c .
From these two equations, eliminating either tang C or tang c, we have either
tang c =
sin A sin a − sin B sin b
sin A cos B cos a − cos A sin B cos b
tang C =
sin A sin a − sin B sin b
,
cos B cos a sin b − cos A sin a cos b
or
to which must be added this equation
sin A sin b = sin B sin a .
Corollary 1
64. Since sin A : sin B = sin a : sin b, we shall have also
tang c =
sin2 a − sin2 b
cos B sin a cos a − cos A sin b cos b
41
and
tang C =
sin2 A − sin2 B
.
sin B cos B sin a − sin A cos A cos b
Corollary 2
65. But paragraphs 43, 44, 53, and 54 again furnish us more convenient
solutions, which are these:
cos 12 (A + B)
sin 12 (A + B)
1
1
1
tang c =
tang
(a
+
b)
=
tang (a − b)
1
1
2
2
2
cos 2 (A − B)
sin 2 (A − B)
and
cos 12 (a − b)
sin 12 (a − b)
1
1
1
tang C =
cot (A + B) =
cot (A − B) ,
1
1
2
2
2
cos 2 (a + b)
sin 2 (a + b)
to which the use of logarithms can easily be applied.
Problem 94
66. Find (Fig. 3) the area of an arbitrary spherical triangle.
Solution
Let EOM be the spherical triangle under consideration; let us name as above
(paragraph 17) the side OE = a, the angle OEM = α, the angle EOM =
y, side OM = x and the angle OM E = φ. This being set, the trilinear
figure M Om will represent the differential of the area which we seek, and
since mn = dx and M n = dy sin x, the product dy dx sin x expresses the
differential of MOm, whence
Z
M Om = dy
dx sin x = dy(1 − cos x)
4
In both the original publication and the Opera Omnia, a second ”Problem 9” occurs
at this point in the text[tr ]
42
and hence the area
Z
EOM = y −
dy cos x .
Now we have found
dy =
C dx
√
,
sin x x2 − CC
so that
C dx cos x
p
.
sin x sin2 x − CC
Z
the area EOM = y −
C
, because C = sin a sin α and since
sin x
p
sin2 x − CC
cos φ =
,
sin x
Then having found sin φ =
we shall have
dφ cos φ = −C dx
cos x
,
sin2 x
thus
C dx cos x
p
dφ = −
sin x sin2 x − CC
Z
and
−
C dx cos x
p
= φ + Const.
sin x sin2 x − CC
Consequently, the area of the triangle EOM will be
= y + φ + Const. = α + y + φ − Const.
To know this constant, suppose y = 0 and since φ then becomes = 180◦ − α,
the area of this vanishing triangle will be = 180◦ −Const. and hence Const. =
180◦ . So we have
the area of the triangle EOM = α + y + φ − 180◦ .
43
Corollary 1
67. Therefore, to find the area of any spherical triangle, one only needs to
take the excess of the sum of its three angles over two right angles, when
the radius of the sphere is expressed by 1. Now in an arbitrary sphere one
will take a great circle arc which is the measure of the said excess, and the
product of this arc by the radius of the sphere will give the desired area of
the spherical triangle.
Corollary 2
68. Thus, the larger a spherical triangle, the more will the sum of its angles
exceed two right angles, and when the area of the triangle occupies one-eighth
of the surface of the sphere, this excess will equal one right angle. For an
arc of a great circle of 90◦ multiplied by the radius gives half the area of the
great circle, hence the eighth part of the surface of the sphere. From this one
will obtain this rule to find the area of the entire spherical triangle. One will
say, as 8 right angles, or 720◦ , is to the excess of the sum of the three angles
over two right angles, so is the whole surface of the sphere to the area of the
triangle under consideration.
44